Sum of absolute differences in a sorted array

You are given an integer array which is  sorted in non-decreasing order. Build and return an integer array result with the same length as the integer array  such that result[i] is equal to the summation of absolute differences between nums[i] and all the other elements in the array.

Example 1 :
Input : nums = [2,3,5]
Output : [4,3,5]
Explanation : Assuming the arrays are 0-indexed, then

result[0] = |2-2| + |2-3| + |2-5| = 0 + 1 + 3 = 4,

result[1] = |3-2| + |3-3| + |3-5| = 1 + 0 + 2 = 3,

result[2] = |5-2| + |5-3| + |5-5| = 3 + 2 + 0 = 5.

Example 2 :
Input : nums = [1,4,6,8,10]
Output : [24,15,13,15,21]

Approach :

1. Take a variable n which is of length of nums
2. Take a variable s which is the sum of nums
3. Create and initialize the result array with all  0’s
4. Create a variable left and initialize it with 0
5. Now traverse the anums array with i as index and num as values in it.
6. Create a variable right and update it as s-left-num
7. Now update result value as nums*(2*i-n+1)+right-left
8. Subsequently keep on adding the num to the left element
9. Return the final result

Time Complexity : O(N)
Space Complexity : O(N)

Below is the implementation:

def getSumAbsoluteDifferences(nums):
    n = len(nums) # take a variable n which is of length of nums
    s = sum(nums) # take a variable s which is the sum of nums

    result = [0] * n # create and initialize the result array with all  0's 
    left = 0 # create a variable left and initialize it with 0
    for i, num in enumerate(nums): # now traverse the anums array with i as index and num as values in it.
        right = s - left - num # create a variavle right and update it as s-left-num
        result[i] = -left + right + num * (2*i - n + 1) # now update result value as nums*(2*i-n+1)+right-left
        left += num # subsequently keep on adding the num to the left element 

    return result #return the fianl result
  
  
nums = [2,3,5]
print(getSumAbsoluteDifferences(nums))

[4, 3, 5]